Question

I need to find dy/dx of the function in the first photo, the second photo is the result i'm supposed to get and the last photo is my result.

Answer 1

I've attached the solution.

Answer 2

Step-by-step explanation:

We begin with the function: $4x^2y-x+y=1-sin(y^2)$

The first thing that we need to do is implicitly differentiate this function with respect to x.

Let us differentiate each piece individually and then combine them.

First, $4x^2y$

There are two thing that we need to keep in mind with this: we will need to apply a product rule and when we differentiate a y with respect to x, we are left with a $\frac{dy}{dx}$.

$4x^2y\\4[(2xy)+(x^2\frac{dy}{dx})}]\\8xy+4x^2\frac{dy}{dx}$

Now, $-x\\-1$

And now, $y\\\\\frac{dy}{dx}$

And lastly, $1-sin(y^2)$

For this, we will need to apply the chain rule.

$1-sin(y^2)\\-(2y\frac{dy}{dx}) cos(y^2)\\-2y\frac{dy}{dx} cos(y^2)$

Now that we have differentiated each part of this function, we can combine them together to get $8xy+4x^2\frac{dy}{dx} -1+\frac{dy}{dx} =-2y\frac{dy}{dx} cos(y^2)$

Our next course of action will be getting all of the terms with $\frac{dy}{dx}$ onto one side so that we can factor it out.

When this is done, we end up with $4x^2\frac{dy}{dx} +\frac{dy}{dx} +2y\frac{dy}{dx} cos(y^2)=-8xy+1$

Once factored, we will get $\frac{dy}{dx} (4x^2+1+2ycos(y^2))=-8xy+1\\\\\frac{dy}{dx}=\frac{-8xy+1}{ 4x^2+1+2ycos(y^2)}\\\\\frac{dy}{dx}=-\frac{8xy-1}{ 4x^2+1+2ycos(y^2)}$

We then were able to isolate $\frac{dy}{dx}$ and then factor out a -1 from the numerator to get our final answer.

Regarding your answer, From what I can see, you ended up incorrectly factoring out the $\frac{dy}{dx}$ from each side before you moved all of the terms to one side. This does not work as there are terms that do not have $\frac{dy}{dx}$ that are having it factored out of them.